I currently do not have enough computing power, so please pardon me for my question, which occurred just recently to me.
So here it goes:
Is the Descartes spoof $$\mathscr{D} = {3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}\cdot{22021} = 198585576189$$ a member of OEIS sequence A228059?
There is an existing Mathematica code in the OEIS hyperlink to test this. Thanks!
The answer to my question is NO, since the abundancy index $I(x):=\sigma(x)/x$ (where $\sigma(x)$ is the sum of the divisors of $x \in \mathbb{N}$) of the first $9$ terms of OEIS sequence A228059 are:
$$I(45) = \frac{26}{15} \approx 1.73333$$ WolframAlpha computation here $$I(405) = \frac{242}{135} \approx 1.79259$$ WolframAlpha computation here $$I(2205) = \frac{494}{245} \approx 2.01633$$ WolframAlpha computation here $$I(26325) = \frac{52514}{26325} \approx 1.99483$$ WolframAlpha computation here $$I(236925) = \frac{474362}{236925} \approx 2.00216$$ WolframAlpha computation here $$I(1380825) = \frac{307086}{153425} \approx 2.00154$$ WolframAlpha computation here $$I(1660725) = \frac{3323138}{1660725} \approx 2.00102$$ WolframAlpha computation here $$I(35698725) = \frac{71396534}{35698725} \approx 1.99997$$ WolframAlpha computation here $$I(3138290325) = \frac{77488034}{38744325} \approx 1.99998$$ WolframAlpha computation here
Notice that, by the definition of OEIS sequence A228059, $|I(x_i)-2|$ must be a (strictly?) decreasing sequence.
Therefore, since $$I(198585576189) = \frac{23622}{11011} \approx 2.14531,$$ it follows that the Descartes spoof $$\mathscr{D} = 198585576189$$ is not a member of OEIS sequence A228059.
Added August 15 2018
In an e-mail correspondence, Tony D. Noe (author of OEIS sequence A228059) says that "(he) found the next term for this sequence: $29891138805 = {5}\cdot({3^2}\cdot{{11}^2}\cdot{71})^2$, (and that) (i)t took $5$ days on a fairly fast Mac."